CHEM1101 2013-J-12 June 2013 In concentration cells no net chemical conversion occurs, however a measurable voltage is present between the two half-cells. Explain how the voltage is produced. 2 In concentration cells, the same solution and electrode are present in both half cells but with different concentrations. For example, both half cells could contain a copper electrode in a copper(ii) sulfate solution but these solutions have different concentrations. In order to equalise the concentrations: In the cell with the lower concentration, oxidation occurs to increase the concentration of the Cu 2+ (aq) in solution, In the other cell, reduction occurs to decrease the concentration of ions in solution. This requires the transfer of electrons between the two half-cells occurs to try and equalise the concentrations in each half-cell and hence to maximise the entropy. Is H 2 a stronger reducing agent under acidic or basic conditions? Give reasons for your answer. 2 When H 2 acts as a reducing agent, it is itself oxidised: H 2 (g) 2H + (aq) + 2e Under acidic conditions, this equilibrium will be pushed to the left due to Le Chatelier s principle. Similarly, it will be pulled to the right under basic conditions as the H + ions produced will react with the OH ions. Therefore H 2 is a better reducing agent under basic conditions.
CHEM1101 2013-J-13 June 2013 A galvanic cell utilises the following redox reaction. NH 4 + (aq) + 8Ce 4+ (aq) + 3H 2 O(l) NO 3 (aq) + 8Ce 3+ (aq) + 10H + (aq) 7 What species is the reducing agent in this reaction? NH 4 + (aq) How many electrons are transferred in the redox reaction? 8 Calculate the standard cell potential, E cell, for this electrochemical cell. From the reduction potential table, E cell o (NO 3 (aq) + 10H + (aq) + 8e E cell o (Ce 4+ (aq) + e Ce 3+ (aq)) = +1.72 V NH 4 + (aq) + 3H 2 O) = +0.88 V The former is reversed and becomes the oxidation half cell: E cell o = (-0.88 + 1.72) V = 0.84 V Calculate ΔG for the redox reaction at 25 C. Using ΔG = -nfe o Answer: +0.84 V ΔG = -(8) (96485 C mol -1 ) (0.84 V) = -650 kj mol -1 Answer: -650 kj mol -1 What is the effect on the E cell of decreasing the concentration of NO 3 (aq) in the anode compartment? If [NO 3 - (aq)] is reduced, the reaction will shift to increase it (Le Chatelier s reaction). The reaction will shift towards products and E cell will increase. Calculate the cell potential, E cell, when [NH 4 + ] = 0.35 M, [Ce 4+ ] = 0.25 M, [NO 3 ] = 5.0 10 2 M, [Ce 3+ ] = 6.0 10 2 M, and the ph is 2.0. As ph = -log[h + (aq)], a ph of 2.0 corresponds to [H + (aq)] = 10-2.0 M. Using the Nernst equation, E cell = E - RT nf lnq = E - RT nf ln NO! 3 aq Ce 3! aq 8 [H! aq ] 10! NH 4 aq Ce 4! aq 8 = (0.84 V) (8.314 J K!1 mol!1 )(298 K) 8(96485 C mol!1 ln (5.0 10!2 )(6.0 10!2 ) 8 (10!2.0 ) ) (0.35)(0.25) 8 = +1.03 V Answer: +1.03 V
CHEM1101 2012-J-14 June 2012 A voltaic cell utilises the following redox reaction. Bi(s) + 3Fe 3+ (aq) + H 2 O(l) 3Fe 2+ (aq) + BiO + (aq) +2H + (aq) 6 What species is the oxidising agent in this reaction? Fe 3+ (aq) How many electrons are transferred in the redox reaction? 3 Calculate the standard cell potential, E cell, for this electrochemical cell. From the data page, the reduction potentials are Fe 3+ (aq) + e - à Fe 2+ (aq) BiO + (aq) + 2H + (aq) + 3e - à Bi(s) + H 2 O(l) E o = +0.77 V E o = +0.32 V The latter is less positive so is reversed to become the oxidation reaction: E o = E o reduction + E o oxidation = (+0.77 V) + (-0.32 V) = +0.45 V Answer: +0.45 V Calculate the equilibrium constant for the redox reaction at 25 C. The equilibrium constant for this 3 electron reaction can be calculated using: E cell = RT nf lnk 0.45 V = (8.314 J K!1 mol!1 )(298 K) (3)(96485 C mol!1 lnk ) lnk = 52.6 so K = 6.8 10 22 Answer: 6.8 10 22 What is the effect on the E cell of decreasing the concentration of BiO + (aq) in the anode compartment? BiO + is a product. Decreasing its concentration will favour formation of more product. From the Nernst equation, shown below, if [BiO + (aq)] is decreased then E cell will increase (i.e. become more positive). E cell = E - RT RT lnq = E - nf 3F ln Fe2! aq 3 BiO! aq [H! aq ] 2 Fe 3! aq 3 ANSWER CONTINUES ON THE NEXT PAGE
CHEM1101 2012-J-14 June 2012 Calculate the cell potential, E cell, when [Fe 3+ ] = 8.2 10 2 M, [Fe 2+ ] = 0.45 M, [BiO + ] = 0.85 M, and the ph is 2.15. As ph = -log 10 [H + (aq)] [H + (aq)] = 10-2.15 From the Nernst equation, E cell = E - RT RT lnq = E - nf 3F ln Fe2! aq = (+0.45 V) (8.314 J K!1 mol!1 )(298 K) (3)(96485 C mol!1 ) = +0.49 V 3 BiO! aq [H! aq ] 2 Fe 3! aq 3 Answer: 0.49 V ln 0.45 3 0.85 [10!2.15] 2 8.2 10!2 3
CHEM1101 2010-J-13 June 2010 The net reaction discharging the lead acid storage battery is: What reaction occurs at the cathode? PbO 2 + Pb + 2H 2 SO 4 2PbSO 4 + 2H 2 O 7 PbO 2 + 2H 2 SO 4 + 2e PbSO 4 + 2H 2 O + SO 4 2 What reaction occurs at the anode? Pb + H 2 SO 4 PbSO 4 + 2H + + 2e Why are the cathode and anode not in separate compartments, as in the Cu/Zn battery? All the redox active species are solids so can be kept apart without needing separate compartments. How does H 2 SO 4 serve as the salt bridge? Which ions flow in which direction to maintain electroneutrality? H 2 SO 4 dissociates to give H +, HSO 4 and SO 4 2 ions which carries the current. The H + ions migrate to the cathode, whilst the HSO 4 and SO 4 2 ions migrate to the anode. What is the formula for the equilibrium constant for the discharge reaction above? Solids do not appear in the equilibrium constant, as their concentration is constant. Water is present as the solvent and is at very high and effectively constant concentration. Hence, only [H 2 SO 4 ] varies: K = [H 2 SO 4 ] 2 The cell potential for this battery is 2.05 V. If the concentration of the H 2 SO 4 is 4.5 M, what is the standard potential of the cell at 25 C? The Nernst equation, E cell = E - lnq, can be used to work out the standard potential for this two electron process. As E cell = 2.05 V when [H 2 SO 4 ] = 4.5 M: 2.05 V = E -. ln(4.5) E = 2.01 V Answer: +2.01 V
CHEM1101 2010-N-13 November 2010 For each electrochemical cell described, write the half-reaction that occurs at each electrode and the overall balanced redox reaction. A voltaic cell constructed using a scandium rod in a solution of scandium(iii) ions (Sc 3+ /Sc) as one half-cell and a nickel rod in a solution of nickel(ii) ions (Ni 2+ /Ni) as the other half-cell. 4 Cathode Ni 2+ + 2e Ni(s) Anode Sc(s) Sc 3+ + 3e Overall cell reaction 3Ni 2+ + 2Sc(s) 3Ni(s) + 2Sc 3+ A voltaic cell in which oxidation of Cr to Cr 3+ by O 2 in the presence of acid occurs. Cathode O 2 (g) + 4H + + 4e H 2 O(l) Anode Cr(s) Cr 3+ + 3e Overall cell reaction 3O 2 (g) + 12H + + 4Cr(s) 6H 2 O(l) + 4Cr 3+ An alkaline battery consists of a powdered Zn/gel anode and a C/MnO 2 cathode. At the anode, Zn is oxidised to Zn 2+ which reacts with the OH ion present in the paste to form Zn(OH) 2 (s). Suppose that an alkaline battery was manufactured using Fe metal instead of Zn metal, and that the Fe was oxidised to Fe 2+ at the anode. What effect would this have on the cell potential or emf of the battery? Explain your answer briefly. 2 The reduction potentials of the Zn 2+ (aq) / Zn and Fe 2+ (aq) / Fe electrodes are: Zn 2+ + 2e Zn(s) E = -0.76 V Fe 2+ + 2e Fe(s) E = -0.44 V This electrode is acting as the anode where oxidation occurs, and cell potential of the battery is: E cell = E ox + E red. Replacing the Zn 2+ /Zn electrode with a Fe 2+ /Fe electrode would reduce the emf of the battery by about 0.32 V (assuming standard concentrations).
CHEM1101 2008-J-14 June 2008 Impure copper can be purified by electrolysis, with the impure copper as one electrode and the purified copper as the other. Is the impure copper the cathode or the anode in the electrolysis cell? 6 The impure copper is the anode. As oxidation occurs at the anode, the impure copper is oxidized to give Cu 2+, leaving the impurities behind. The Cu 2+ is then reduced at the cathode to pure Cu(s). If a battery is used as the power source, is the positive terminal of the battery connected to the impure copper or to the pure copper electrode? In an electrolytic cell, oxidation at the anode must be forced to occur. This is achieved by making the anode positive so that it removes electrons from the reactant (impure copper) at the electrode. If electrolysis for 1.0 hour with a current of 5.2 A produces 5.9 g of pure copper, calculate the oxidation number of the copper dissolved in the cell. As the atomic mass of copper is 63.55 g mol -1, 5.9 g corresponds to: number of moles of copper = mass 5.9 g = = 0.093 mol atomic mass 63.55 mol -1 A current, I, of 5.2 A applied for a time, t, of 1.0 hour corresponds to: number of moles of electons = It F (5.2 A) (1.0 60 60 s) = = 0.19 mol (96485 C mol -1 ) Each mole of copper therefore requires 0.19 2.09. As the oxidation number 0.093 is a whole number, this corresponds to an oxidation number of +2. Oxidation number: +2 (II) Explain, with the use of standard reduction potentials, why a silver impurity in the copper can be recovered from the cell as silver metal, but a nickel impurity is found dissolved in the electrolyte solution. The relevant reduction potentials are +0.34 V for Cu 2+ (aq), +0.80 V for Ag + (aq) and -0.24 V for Ni 2+ (aq). ANSWER CONTINUES ON THE NEXT PAGE
CHEM1101 2008-J-14 June 2008 For formation of silver metal, Cu(s) Cu 2+ (aq) + 2e - (aq) Ag + (aq) + e - (aq) 2Ag(s) E = -0.34 V (reversed for oxidation) E = +0.80 V A voltage of 0.34 V is sufficient to dissolve copper but not to dissolve silver. The silver thus remains as solid Ag(s). The reaction of Ag(s) with Cu 2+ (aq) is unfavourable as the reverse reaction is favourable: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) E = ((-0.34) + (+0.80)) V = +0.46 V As E > 0, the reaction should occur. Silver metal can thus be formed in the presence of copper metal. For formation of nickel metal, Cu(s) Cu 2+ (aq) + 2e - (aq) Ni 2+ (aq) + 2e - (aq) Ni(s) E = -0.34 V (reversed for oxidation) E = -0.24 V The overall reaction is: Cu(s) + Ni 2+ (aq) Cu 2+ (aq) + Ni(s) E = ((-0.34) + (-0.24)) V = -0.58 V As E < 0, the reaction will not occur. Nickel metal cannot be formed from Ni 2+ in the presence of copper metal. Explain what happens to an iron impurity in the Cu. The reduction potential for Fe 2+ (aq) is -0.44 V. Thus, for formation of iron metal, Cu(s) Cu 2+ (aq) + 2e - (aq) Fe 2+ (aq) + 2e - (aq) Fe(s) E = -0.34 V (reversed for oxidation) E = -0.44 V The overall reaction is: Cu(s) + Fe 2+ (aq) Cu 2+ (aq) + Fe(s) E = ((-0.34) + (-0.44)) V = -0.78 V As E < 0, the reaction will not occur. Iron metal cannot be formed from Fe 2+ in the presence of copper metal. The iron impurity will stay in the electrolyte solution as Fe 2+.
CHEM1101 2006-N-12 November 2006 Lead sulfate is used as a white pigment and also in car batteries. Its solubility in water at 25 C is 4.25 10 3 g per100 ml of solution. Write an equation for the dissolution of lead sulfate in water and determine K sp at 25 C. 3 The dissolution equilibrium is: PbSO 4 (s) Pb 2+ (aq) + SO 4 2- (aq) with K sp = [Pb 2+ (aq)][so 4 2- (aq)] The molar mass of PbSO 4 is (207.2 (Pb)) + (32.07 (S)) + (4 16.00 (O)) = 303.27 so 4.25 10-3 g corresponds to: moles of PbSO 4 = -3 4.25 10 =1.40 10-5 moles 303.27 As this amount dissolves in 100 ml, 1.40 10-4 moles will dissolve in a litre. From the dissolution equilibrium, 1 mole of PbSO 4 gives rise to 1 mole of Pb 2+ (aq) and 1 mole of SO 4 2- (aq) so K sp is given by: K sp = [Pb 2+ (aq)][so 4 2- (aq)] = (1.40 10-4 ) (1.40 10-4 ) = 1.96 10-8 K sp = 1.96 10-8 A voltaic cell consists of Ni/Ni 2+ and Co/Co 2+ half cells with initial concentrations of [Ni 2+ ] = 0.80 M and [Co 2+ ] = 0.20 M. What is the initial E cell at 298 K? 5 The standard reduction potentials and half cells are: Co 2+ (aq) + 2e - Co(s) Ni 2+ (aq) + 2e - Ni(s) ) E 0 = 0.28 V E 0 = -0.24 V The least positive (Co/Co 2+ ) cell is reversed giving the overall cell reaction as: Co(s) + Ni 2+ (aq) Co 2+ (aq) + Ni(s) E 0 = (+0.28) + (-0.24) = +0.04 V This is the cell potential will all reactants and products in their standard (1M) concentrations. At other concentrations, the Nernst equation gives the potential as: 0 2.303RT E cell = E - log(q) nf The overall reaction is a 2e - 2+ [Co (aq)] process (n = 2) with Q = 2+ [Ni (aq)], hence, 2+ 0 2.303RT [Co (aq)] (2.303) (8.314) (298) 0.20 E cell = E - log =(+0.04)- log = +0.06 V 2F 2+ [Ni (aq)] 2 96485 0.80 E cell = +0.06 V
CHEM1101 2006-N-12 November 2006 What is the value of the equilibrium constant, K, for this cell? The equilibrium constant is related to E 0 by: 0 2.303RT E = log(k) nf so log(k) = 0 E nf 2.303RT = (+0.04) (2) (96485) =1.35 (2.303) (8.314) (298) Hence: 1.35 K=10 =23 K = 23